Question 3.(10 points)A Web ad can be designed from three different colors, thre

Question

Question 3.(10 points)A Web ad can be designed from three different colors, three font types, three font sizes, three images, and five text phrases. A specific design is randomly generated by the Web server when you visit the site. Let A denote the event that the design color is red, and let B denote the event that the font size is not the smallest one. Use the addition rules to calculate the following probabilities.(a) P(AB) (b) P(AB) (c) P(A B)

Question 4.(10 points)Consider the well failure data from Question 2.(a) What is the probability of a failure given there are more than 1,000 wells in a geological formation? (b) What is the probability of a failure given there are fewer than 500 wells in a geological formation?

Question 5.(20 points)A lot of 100 semiconductor chips contains 10 that are defective.(a) Two are selected, at random, without replacement, from the lot. Determine the probability that the second chip selected is defective.(b) Three are selected, at random, without replacement, from the lot. Determine the probability that all are defective.

Question 6.(25 points)The probability that a lab specimen contains high levels of contamination is 0.10. Four samples are checked, and the samples are independent.(a) What is the probability that none contain high levels of contamination?(b) What is the probability that exactly one contains high levels of contamination?(c) What is the probability that at least one contains high levels of contamination?

Explanation / Answer

Question 10

Here A = Design color is Red

Pr(A) = 1/3

B = Font size is not the smallest ne = 1 - 1/3 = 2/3

P(A U B) = Pr( design color is Red or Font sizze is not the smallest one) = 1/3 + 2/3 - 1/3 * 2/3 = 7/9

P(A U B') = Pr(Degin color is red or font size is smallest one) = 1/3 + 1/3 - 1/3 * 1/3 = 5/9

P(A' U B') = Pr(Design color is not red or font size is smallest one) = 2/3 + 1/3 - 2/3 * 1/3 = 7/9

Question 4

Question 2 is not given here.

Question 5

(a) Pr(Second chip is defective) = Pr(Second chip is defective) = Pr(First chip is defective) * Pr(Second chip is defective) + Pr(First chip is not defective) * Pr(Second chip is defective) = 90/100 * 10/99 + 10/100 * 9/99 = 0.1

(b) Pr(All three are defective) = (10/100) * (9/99) * (8/98) = 0.00074

Question 6

Pr(High levels of contamination) = 0.10

(a) If X is the number of contaminated samples out of 4.

Pr(X = 0 ; 4 ; 0.10) = 4C0 (0.10)0(0.90)4 = 0.6561

(b) Pr( X = 1; 4 ; 0.10) = 4C1 (0.10)1(0.90)3 = 0.2916

(c) Pr(X >=1 ; 4 ; 0.1) = 1 - Pr(X = 0) = 1 - 0.6561 = 0.3439

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